Discriminant of algebraic number field is $0$ or $1 \bmod 4$

Question

I am reading the following theorem of Ludwig Stickelberger in Introductory Algebraic Number Theory written by S. Alaca and Kenneth S. Williams

Let $K$ be an algebraic number field of degree $n$. Then the discriminant $d(K)$ of $K$ satisfies $d(K) \equiv 0 \text{ or } 1 \bmod 4$

To show this,
Let $\{w_1,w_2, \ldots,w_n \} $ be an integral basis for $K$. Let $\{w_i^{(j)} \, \vert \, j = 1,2 , \ldots n \}$ be the set of $K$ - conjugates of $w_i$ with $w_i^{(1)} = w_i$ for $i = 1,2 , \ldots ,n$ .

A fairly simple induction proof is sufficient to show that the expansion of determinant of $\det = \begin{vmatrix} w_1^{(1)} && w_2^{(1)} && \cdots && w_n ^{(1)} \\ \vdots && \cdots && \cdots &&\vdots\\ w_n^{(1)} && w_2^{(n)} && \cdots && w_n^{(n)} \end{vmatrix}$

contains $n!$ terms, half of which occur with positive signs and half with negative signs.

Let the sum of those with positive signs be $\lambda$ and those with negative signs $\mu$ so that $$ \det = \lambda - \mu$$

Let $A = \lambda + \mu$ and $B= \lambda \mu$, then $d(K)= \det^2 = (\lambda - \mu)^2 = A^2 - 4B$

It is easy to see that $w_i^{(j)}$ is an algebraic integer and so does $\lambda$ and $\mu$ since the set of algebraic integers ($\Omega$) form integral domain. For the same reason, we see that $A, B \in \Omega$.

Now the aim is to show that $A$ and $B$ are rational numbers. We note that there exist $\theta \in \Omega$ such that $K = \mathbb{Q}(\theta)$ and let $\theta_1 = \theta, \theta_2, \ldots ,\theta_n$ be conjugates of $\theta$ over rational numbers. Then the author says

If we express each $w_j$ as a polynomial in $\theta$ with rational coefficients, $A$ becomes a symmetric function of $\theta_1, \cdots , \theta_n$ with rational coefficients and so $A \in \mathbb{Q}$

I know the author is using the fundamental theorem of symmetric polynomials to conclude that $A$ is a rational number. But I have issues with the above claim.

$1$. I think it should be

$\dots w_j$ as a polynomial in $\theta_j$ with rational coefficients $\dots$

$2$. I do not understand how $A$ is a symmetric function in terms of $\theta_1, \ldots, \theta_n$. Of course, it is a function of $\theta_i$s but how we can conclude that it is a symmetric function? When I investigate the cases $n=2,3$ it is apparent. But it is becoming tedious to do for higher values of $n$.

What is the missing piece of the puzzle?

Answer 1

Taking hints from above comments, using the definition of determinant given here, we see that $$\det = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{i=1}^n w_{\sigma(i)}^{(i)} = \sum_{\sigma \in A_n} \prod_{i=1}^n w_{\sigma(i)}^{(i)} - \sum_{\sigma \not \in A_n} \prod_{i=1}^n w_{\sigma(i)}^{(i)} $$

So we take $\lambda = \sum_{\sigma \in A_n} \prod_{i=1}^n w_{\sigma(i)}^{(i)} $ and $\mu = \sum_{\sigma \not \in A_n} \prod_{i=1}^n w_{\sigma(i)}^{(i)} $

Let $w_i = f_i(\theta)$ where $f_i$ is a polynomial with rational coefficients then $w_{\sigma(i)}^{(i)} = f_{\sigma(i)}(\theta^{(i)}) = f_{\sigma(i)}(\theta_{i})$. So

$$\lambda(\theta_1,\ldots\theta_n ) = \sum_{\sigma \in A_n} \prod_{i=1}^n f_{\sigma(i)}(\theta_{i}) $$ and
$$\mu(\theta_1,\ldots\theta_n ) = \sum_{\sigma \not \in A_n} \prod_{i=1}^n f_{\sigma(i)}(\theta_{i}) $$

Let $\alpha \in A_n$, then $$ \lambda(\theta_{\alpha(1)},\ldots,\theta_{\alpha(n)}) =\sum_{\sigma \in A_n} \prod_{i=1}^n f_{\sigma(\alpha(i))}(\theta_{\alpha(i)}) = \sum_{\sigma \circ \alpha \in A_n} \prod_{i=1}^n f_{\sigma(\alpha(i))}(\theta_{\alpha(i)}) = \lambda $$ and similarly $$ \mu(\theta_{\alpha(1)},\ldots,\theta_{\alpha(n)}) =\sum_{\sigma \not\in A_n} \prod_{i=1}^n f_{\sigma(\alpha(i))}(\theta_{\alpha(i)}) = \sum_{\sigma \circ \alpha \not \in A_n} \prod_{i=1}^n f_{\sigma(\alpha(i))}(\theta_{\alpha(i)}) = \mu$$

So it shows that $\lambda $ and $\mu$ is invariant under even permutations.

In a similar manner it is easy to see that for any $\alpha \not\in A_n$, $$\lambda(\theta_{\alpha(1)},\ldots,\theta_{\alpha(n)}) = \mu$$ and $$\mu(\theta_{\alpha(1)},\ldots,\theta_{\alpha(n)}) = \lambda$$

So $A= A(\theta_1, \ldots , \theta_n) = \lambda + \mu $ is invariant under both odd and even permutations hence symmetric polynomial in terms of $\theta_i$. So using fundamental theorem of symmetric polynomials we can say that $A\in \mathbb{Q}$ as values of elementary symmetric polynomials are rational numbers. Also $B = \frac14(A^2 - d(K)) \in \mathbb{Q}$.

To finish the proof we see that $A, B\in \Omega$ so that $A , B\in \mathbb{Z}$ which shows that $d(K) \equiv 0 , 1 \mod4$.

This answers my both questions,

1. We can express $w_j$ in terms of $\theta$ and finish the proof.
2. $A$ is a symmetric polynomial in terms of $\theta_i.$